How do you find the derivative of #y= arctan[(x^2-1)^(1/2)] + "arccsc"(x)#? Calculus Differentiating Trigonometric Functions Differentiating Inverse Trigonometric Functions 1 Answer Sasha P. Sep 15, 2015 0 Explanation: #y'=1/(((x^2-1)^(1/2))^2+1)1/2(x^2-1)^(-1/2)2x+(arcsin(1/x))'# #=1/(x^2-1+1)1/2(2x)/sqrt(x^2-1)+1/sqrt(1-(1/x)^2)(-1/x^2)=# #=x/(x^2sqrt(x^2-1))-1/(x^2sqrt((x^2-1)/x^2))=# #=1/(xsqrt(x^2-1))-1/(xsqrt(x^2-1))=0# Answer link Related questions What is the derivative of #f(x)=sin^-1(x)# ? What is the derivative of #f(x)=cos^-1(x)# ? What is the derivative of #f(x)=tan^-1(x)# ? What is the derivative of #f(x)=sec^-1(x)# ? What is the derivative of #f(x)=csc^-1(x)# ? What is the derivative of #f(x)=cot^-1(x)# ? What is the derivative of #f(x)=(cos^-1(x))/x# ? What is the derivative of #f(x)=tan^-1(e^x)# ? What is the derivative of #f(x)=cos^-1(x^3)# ? What is the derivative of #f(x)=ln(sin^-1(x))# ? See all questions in Differentiating Inverse Trigonometric Functions Impact of this question 1916 views around the world You can reuse this answer Creative Commons License